dr. alagoz chapter 2 analysis of algorithms algorithm input output an algorithm is a step-by-step...

Dr. Alagoz

CHAPTER 2Analysis of Algorithms

AlgorithmInput Output

An algorithm is a step-by-step procedure forsolving a problem in a finite amount of time.

Analysis of Algorithms 2

Dr. AlagozAlgorithm Analysis: Do NOT worry !!!


Dr. Alagoz

properties of logarithms:logb(xy) = logbx + logby

logb (x/y) = logbx - logby

logbxa = alogbx

logba = logxa/logxbproperties of exponentials:a(b+c) = aba c

abc = (ab)c

ab /ac = a(b-c)

b = a logab

bc = a c*logab

Series summationsLogarithms and Exponents

Proof techniquesBasic probability

Math you need to Review


Dr. Alagoz

Assumptions for the computational model

Basic computer with sequentially executed instructions

Infinite memory

Has standard operations; addition, multiplication, comparison in 1 time unit unless stated otherwise

Has fixed-size (32bits) integers such that no-fancy operations are allowed!!! eg.matrix inversion,

Proof techniquesBasic probability


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Running Time Most algorithms transform input objects into output objects.The running time of an algorithm typically grows with the input size.Average case time is often difficult to determine.We focus on the worst case running time.

Easier to analyze Crucial to applications

such as games, finance, image,robotics, AI, etc.

0

20

40

60

80

100

120

Runnin

g T

ime

1000 2000 3000 4000

Input Size

best caseaverage caseworst case


Dr. Alagoz

Experimental Studies

Write a program implementing the algorithmRun the program with inputs of varying size and compositionAssume a subprogram is written to get an accurate measure of the actual running timePlot the results 0

1000

2000

3000

4000

5000

6000

7000

8000

9000

0 50 100

Input Size

Tim

e (

ms)


Dr. Alagoz

Limitations of Experiments

It is necessary to implement the algorithm, which may be difficultResults may not be indicative of the running time on other inputs not included in the experiment. In order to compare two algorithms, the same hardware and software environments must be used


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Theoretical AnalysisUses a high-level description of the algorithm instead of an implementationCharacterizes running time as a function of the input size, n.Takes into account all possible inputsAllows us to evaluate the speed of an algorithm independent of the hardware/software environment


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Pseudocode

High-level description of an algorithmMore structured than English proseLess detailed than a programPreferred notation for describing algorithmsHides program design issues

Algorithm arrayMax(A, n)Input array A of n integersOutput maximum element of A

currentMax A[0]for i 1 to n 1 do

if A[i] currentMax thencurrentMax A[i]

return currentMax

Example: find max element of an array


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Pseudocode Details

Control flow if … then … [else …] while … do … repeat … until … for … do … Indentation replaces

braces

Method declarationAlgorithm method (arg [, arg…])

Input …Output …

Method callvar.method (arg [, arg…])

Return valuereturn expression

Expressions Assignment

(like in Java) Equality testing

(like in Java)n2 Superscripts and

other mathematical formatting allowed


Dr. Alagoz

Important Functions Seven functions that often appear in algorithm analysis:

Constant 1 Logarithmic log n Linear n N-Log-N n log n Quadratic n2

Cubic n3

Exponential 2n

In a log-log chart, the slope of the line corresponds to the growth rate of the function

1E+01E+21E+41E+61E+8

1E+101E+121E+141E+161E+181E+201E+221E+241E+261E+281E+30

1E+0 1E+2 1E+4 1E+6 1E+8 1E+10n

T(n

)

Cubic

Quadratic

Linear


Dr. Alagoz

Primitive Operations

Basic computations performed by an algorithm

Identifiable in pseudocode

Largely independent from the programming language

Exact definition not important (we will see why later)

Assumed to take a constant amount of time in the RAM model

Examples: Evaluating an

expression Assigning a

value to a variable

Indexing into an array

Calling a method Returning from a

method


Dr. AlagozDetailed Model 4 counting operations-1

fetchto fetch an integer operant from memory

storeto store an integer result in memory

+to add two integers

-to subtract two integers

xto multiply two integers

/to divide two integers

<to comparison of two integers

The time require for the following operations are all constants.


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returnto return from a method

callto call a method

storeto pass an integer argument to a method

[.]for the address calculation (not including the computation of the subscription)

newto allocate a fixed amount of storage from the heap using new

Detailed Model 4 counting operations-2

The time require for the following operations are all constants.


Dr. Alagoz

Statement Time requiredy = x; fetch + store

y = 1; fetch + store

y = y + 1; 2 fetch + + + store

y += 1; 2 fetch + + + store

++y; 2 fetch + + + store

y++; 2 fetch + + + store

Detailed Model 4 counting operations-3


Dr. AlagozDetailed Model... Example1:Sum

...public static int sum(int n) {

int result = 0;for (int i = 1; i <= n; ++i){

result += i;}return result;

}...

Statement Time requiredint result = 0; fetch + store

int i = 1 fetch + store

i <= n (2fetch + < ) (n+1)++i (2fetch + + + store)

nresult += i (2fetch + + + store)

nreturn result fetch + return

n

i

i1


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Detailed Model... Example2

y = a [i]

3fetch + [.] + store

operations fetch a (the base address of the

array) fetch i (the index into the array) address calculation fetch array element a[i] store the result


Dr. AlagozDetailed Model...Example3: Horner (*)

public class Horner {

public static void main(String[] args) {Horner h = new Horner();int[] a = { 1, 3, 5 };System.out.println("a(1)=" + h.horner(a, a.length - 1,

1));System.out.println("a(2)=" + h.horner(a, a.length - 1,

2));}

int horner(int[] a, int n, int x) {int result = a[n];for (int i = n - 1; i >= 0; --i) {

result = result * x + a[i];/**/System.out.println("i=" + i + " result" +

result);}return result;

}}

i=1 result8i=0 result9a(1)=9i=1 result13i=0 result27a(2)=27

outputoutput

in

ii xa

0


Dr. Alagoz

Detailed Model...Example-4 (*)public class FindMaximum {

public static void main(String[] args) {FindMaximum h = new FindMaximum();int[] a = { 1, 3, 5 };System.out.println("max=" + h.findMaximum(a));

}

int findMaximum(int[] a) {int result = a[0];for (int i = 0; i < a.length; ++i) {

if (result < a[i]) {result = a[i];

}System.out.println("i=" + i + " result=" + result);

}return result;

}}

i=0 result=1i=1 result=3i=2 result=5max=5

outputoutput

3fetch + [.] + store

fetch + store(2fetch + <) n(2fetch + + + store) (n-1)

(4fetch + [.] + <) (n-1)

(3fetch + [.] + store) ?

fetch + store


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Simplified Model … More Simplification

All timing parameters are expressed in units of clock cycles. In effect, T=1. The proportionality constant, k, for all timing parameters is assumed to be the same: k=1.


Dr. Alagoz

Simplified Model …Example_FindMax public class FindMaximum {

public static void main(String[] args) {FindMaximum h = new FindMaximum();int[] a = { 1, 3, 5 };System.out.println("max=" + h.findMaximum(a));

}



}System.out.println("i=" + i + " result=" + result);

}return result;

}}

i=0 result=1i=1 result=3i=2 result=5max=5

outout

detailed2 3fetch + [.] + store

3a fetch + store3b (2fetch + <) n3c (2fetch + + + store) (n-1)

4 (4fetch + [.] + <) (n-1)

6 (3fetch + [.] + store) ?

9 fetch + store

simple2 5

3a 2

3b (3)n

3c (4)(n-1)

4 (6)(n-1)

6 (5)?

9 2

12345678910


Dr. AlagozSimplified Model > Algorithm1…Geometric Series Sum

public class GeometrikSeriesSum {

public static void main(String[] args) {System.out.println("1, 4: " + geometricSeriesSum(1, 4));System.out.println("2, 4: " + geometricSeriesSum(2, 4));

}

public static int geometricSeriesSum(int x, int n) {int sum = 0;for (int i = 0; i <= n; ++i) {

int prod = 1;for (int j = 0; j < i; ++j) {

prod *= x;}sum += prod;

}return sum;

}} 1, 4: 5

2, 4: 31

outputoutput

simple12 23a 23b 3(n+2)3c 4(n+1)4 2(n+1)5a 2(n+1)5b 3 (i+1)5c 4 i6 4 i78 4(n+1)910 21112

Total 11/2 n2 + 47/2 n +27

123456789101112

n

i 0

n

i 0

n

i 0


Dr. AlagozSimplified Model > Algorithm_SumHornerGeometric Series Sum …public class GeometricSeriesSumHorner {

public static void main(String[] args) {System.out.println("1, 4: " + geometricSeriesSum(1, 4));System.out.println("2, 4: " + geometricSeriesSum(2, 4));

}

public static int geometricSeriesSum(int x, int n) {int sum = 0;for (int i = 0; i <= n; ++i) {

sum = sum * x + 1;}return sum;

}}

1, 4: 52, 4: 31

outputoutput

2 23a 23b 3(n+2)3c 4(n+1)4 6(n+1)6 2

Total 13 n + 22

1234567


Dr. Alagoz

Simplified Model > Algorithm_SumPower Geometric Series Sum …

public class GeometrikSeriesSumPower {

public static void main(String[] args) { System.out.println("1, 4: " + powerA(1, 4)); System.out.println("1, 4: " + powerB(1, 4)); System.out.println("2, 4: " + powerA(2, 4)); System.out.println("2, 4: " + powerB(2, 4));

}

...public static int powerA(int x, int n) {

int result = 1;for (int i = 1; i <= n; ++i) {

result *= x;}return result;

}

public static int powerB(int x, int n) {if (n == 0) {

return 1;} else if (n % 2 == 0) { // n is even

return powerB(x * x, n / 2);} else { // n is odd

return x * powerB(x * x, n / 2);}

}}

1, 4: 11, 4: 12, 4: 162, 4: 16

outputoutput

powerB 0<n 0<n n=0 n even n odd

10 3 3 311 2 - -12 - 5 513 - 10+T(n/2) -15 - - 12+T(n/2)

Total 5 18+T(n/2) 20+T(n/2)

1234567891011121314151617

powerB 1 n=0

xn = (x2)n/2 0<n, n is even x(x2)n/2 0<n, n is odd

powerB 5 n=0

xn = 18+T(n/2) 0<n, n is even 20+T(n/2) 0<n, n is odd


Dr. AlagozSimplified Model > Calculation of PowerB …Geometric Series Sum … (*)

Let n = 2k for some k>0.Since n is even, n/2 = n/2 = 2k-1.

For n = 2k, T(2k) = 18 + T(2k-1).

Using repeated substitutionT(2k) = 18 + T(2k-1)

= 18 + 18 + T(2k-2) = 18 + 18 + 18 + T(2k-3) …= 18j + T(2k-j)

Substitution stops when k = jT(2k) = 18k + T(1)

= 18k + 20 + T(0)= 18k + 20 + 5= 18k + 25.

n = 2k then k = log2 n

T(2k) = 18 log2 n + 25

powerBxn = 18+T(n/2) 0<n

n is even


Dr. Alagoz

Suppose n = 2k–1 for some k>0.Since n is odd, n/2 = (2k–1)/2 = (2k–2)/2= 2k-1

For n = 2k-1, T(2k-1) = 20 + T(2k-1-1), k>1.

Using repeated substitutionT(2k-1) = 20 + T(2k-1-1)

= 20 + 20 + T(2k-2-1) = 20 + 20 + 20 + T(2k-3-1) …= 20j + T(2k-j-1)

Substitution stops when k = jT(2k-1) = 20k + T(20-1)

= 20k + T(0)= 20k + 5.

n = 2k-1 then k = log2 (n+1)

T(n) = 20 log2 (n+1) + 5

Therefore, powerB 5 n=0

xn = 18+T(n/2) 0<n, n is even 20+T(n/2) 0<n, n is odd

Average: 19(log2(n+1) + 1) + 18

Simplified Model > Calculation of PowerB Geometric Series Sum … (*)

n is odd


Dr. Alagoz

public class GeometrikSeriesSumPower {

public static void main(String[] args) {System.out.println(“s 2, 4: " + geometrikSeriesSumPower

(2, 4));}

...public static int geometrikSeriesSumPower (int x, int n) {

return powerB(x, n + 1) – 1 / (x - 1);}

public static int powerB(int x, int n) {if (n == 0) {

return 1;} else if (n % 2 == 0) { // n is even

return powerB(x * x, n / 2);} else { // n is odd

return x * powerB(x * x, n / 2);}

}}

s 2, 4: 31

outputoutput

Simplified Model > Algorithm_SumPower Geometric Series Sum …


Dr. AlagozComparison of 3 Algorithms

algorithm T(n) Sum 11/2 n2 + 47/2 n + 27Horner 13n + 22Power 19(log2(n+1) + 1) + 18

sum

Horner

Power


Dr. Alagoz

Counting Primitive Operations for

Pseudocodes By inspecting the pseudocode, we can determine the maximum number of primitive operations executed by an algorithm, as a function of the input size

Algorithm arrayMax(A, n)

# operations

currentMax A[0] 2for i 1 to n 1 do 2n

if A[i] currentMax then 2(n 1)currentMax A[i] 2(n 1)

{ increment counter i } 2(n 1)return currentMax 1

Total 8n 2


Dr. Alagoz

Estimating Running Time

Algorithm arrayMax executes 8n 2 primitive operations in the worst case. Define:a = Time taken by the fastest primitive operationb = Time taken by the slowest primitive

operation

Let T(n) be worst-case time of arrayMax. Then

a (8n 2) T(n) b(8n 2)

Hence, the running time T(n) is bounded by two linear functions


Dr. AlagozAsymptotic Algorithm Analysis

The asymptotic analysis of an algorithm determines the running time in big-Oh notationTo perform the asymptotic analysis

We find the worst-case number of primitive operations executed as a function of the input size

We express this function with big-Oh notation

Example: We determine that algorithm arrayMax executes at

most 8n 2 primitive operations We say that algorithm arrayMax “runs in O(n) time”

Since constant factors and lower-order terms are eventually dropped anyhow, we can disregard them when counting primitive operations


Dr. AlagozGrowth Rate of Running Time

Changing the hardware/ software environment Affects T(n) by a constant factor, but Does not alter the growth rate of T(n)

The linear growth rate of the running time T(n) is an intrinsic property of algorithm arrayMax


Dr. Alagoz

Constant Factors

The growth rate is not affected by

constant factors or

lower-order terms

Examples 102n 105 is a

linear function 105n2 108n is a

quadratic function 1E+01E+21E+41E+61E+8

1E+101E+121E+141E+161E+181E+201E+221E+241E+26

1E+0 1E+2 1E+4 1E+6 1E+8 1E+10n

T(n

)

Quadratic

Quadratic

Linear

Linear


Dr. Alagoz

Big-Oh Notation

Given functions f(n) and g(n), we say that f(n) is O(g(n)) if there are positive constantsc and n0 such that

f(n) cg(n) for n n0

Example: 2n 10 is O(n) 2n 10 cn (c 2) n 10 n 10(c 2) Pick c 3 and n0 10

1

10

100

1.000

10.000

1 10 100 1.000n

3n

2n+10

n


Dr. Alagoz

Big-Oh Example

Example: the function n2 is not O(n)

n2 cn n c The above

inequality cannot be satisfied since c must be a constant

1

10

100

1.000

10.000

100.000

1.000.000

1 10 100 1.000n

n 2̂

100n

10n

n


Dr. Alagoz

More Big-Oh Examples

7n-2

7n-2 is O(n)need c > 0 and n0 1 such that 7n-2 c•n for n n0

this is true for c = 7 and n0 = 1

3n3 + 20n2 + 53n3 + 20n2 + 5 is O(n3)need c > 0 and n0 1 such that 3n3 + 20n2 + 5 c•n3 for n

n0

this is true for c = 4 and n0 = 21 3 log n + 53 log n + 5 is O(log n)need c > 0 and n0 1 such that 3 log n + 5 c•log n for n

n0

this is true for c = 8 and n0 = 2


Dr. Alagoz

Big-Oh and Growth RateThe big-Oh notation gives an upper bound on the growth rate of a functionThe statement “f(n) is O(g(n))” means that the growth rate of f(n) is no more than the growth rate of g(n)

We can use the big-Oh notation to rank functions according to their growth rate

f(n) is O(g(n)) g(n) is O(f(n))

g(n) grows more

Yes No

f(n) grows more No Yes

Same growth Yes Yes


Dr. Alagoz

Big-Oh Rules

Use the smallest possible class of functions Say “2n is O(n)” instead of “2n is O(n2)”

Use the simplest expression of the class Say “3n 5 is O(n)” instead of “3n 5 is

O(3n)”

Theorem

Consider polynomial where am>0.

Then f(n) = O(nm). i.e.,1. Drop lower-order terms2. Drop constant factors

m

i

iinanf

0

)(


Dr. Alagoz

Big-Oh Rules-2 logk n = O(n) for any constant kZ+

f(n) = O (f(n))

c O(f(n)) = O (f(n))

O(f(n)) + O(f(n)) = O(f(n))

O(O(f(n))) = O(f(n))

O(f(n)) O(g(n)) = O(f(n) g(n))


Dr. Alagoz

Relatives of Big-Oh

big-Omega f(n) is (g(n)) if there is a constant c > 0

and an integer constant n0 1 such that

f(n) c•g(n) for n n0

big-Theta f(n) is (g(n)) if there are constants c’ > 0

and c’’ > 0 and an integer constant n0 1 such that c’•g(n) f(n) c’’•g(n) for n n0


Dr. Alagoz

Intuition for Asymptotic Notation

Big-Oh f(n) is O(g(n)) if f(n) is

asymptotically less than or equal to g(n)

big-Omega f(n) is (g(n)) if f(n) is

asymptotically greater than or equal to g(n)

big-Theta f(n) is (g(n)) if f(n) is

asymptotically equal to g(n)


Dr. AlagozExample Uses of the Relatives of Big-Oh

f(n) is (g(n)) if it is (n2) and O(n2). We have already seen the former, for the latter recall that f(n) is O(g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) < c•g(n) for n n0

Let c = 5 and n0 = 1

5n2 is (n2)

f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0

let c = 1 and n0 = 1

5n2 is (n)

f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0 1 such that f(n) c•g(n) for n n0

let c = 5 and n0 = 1

5n2 is (n2)


Dr. Alagoz

Comparison of OrdersO(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) <

O(2n)


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O(log n)

Comparison of OrdersO(1) < O(log n) < O(n) < O(n log n) < O(n2) < O(n3) <

O(2n)

O(1)

O(2n)

O(n3)

O(n2)

O(n log n)O(n)


Dr. Alagoz

O(n) Analysis of Running Time

Example Worst-case running time

if statement 5 executes all the time…

Best-case running time if statement 5 never executes…

On-the-average running time if statement 5 executes half of the time????



}}return result;

}

123456789


Dr. AlagozAlgorithm Complexity as a function of size n

Problem Input of size n Basic operation

Searching a list

Sorting a list

Multiplying two matrices

Prime factorization

Evaluating a polynomial

Traversing a tree

Towers of Hanoi

lists with n elements

lists with n elements

two n-by-n matrices

n-digit number

polynomial of degrees n

tree with n nodes

n disks

comparison

comparison

multiplication

division

multiplication

accessing a node

moving a disk

A comparison-based algorithm for searching or sorting a list is

based on

•making comparisons involving list elements

•then making decisions based on these comparisons.

• SOON U ll know all about these!!!


Dr. Alagoz

HWLA

Solve Exercises in Chapter2 : 1, 7, 9 ,12, 25, 29, 31


Dr. AlagozComputing Prefix Averages(*)

We further illustrate asymptotic analysis with two algorithms for prefix averagesThe i-th prefix average of an array X is average of the first (i 1) elements of X:

A[i] X[0] X[1] … X[i])/(i+1)

Computing the array A of prefix averages of another array X has applications to financial analysis such as ARMA, ARIMA, FARIMA, models.

0

5

10

15

20

25

30

35

1 2 3 4 5 6 7

X

A


Dr. Alagoz

Prefix Averages (Quadratic) (*)

The following algorithm computes prefix averages in quadratic time by applying the definition

Algorithm prefixAverages1(X, n)Input array X of n integersOutput array A of prefix averages of X #operations A new array of n integers nfor i 0 to n 1 do n

s X[0] nfor j 1 to i do 1 2 … (n 1)

s s X[j] 1 2 … (n 1)A[i] s (i 1) n

return A 1


Dr. Alagoz

Arithmetic Progression (*)

The running time of prefixAverages1 isO(1 2 …n)The sum of the first n integers is n(n 1) 2

There is a simple visual proof of this fact

Thus, algorithm prefixAverages1 runs in O(n2) time 0

1

2

3

4

5

6

7

1 2 3 4 5 6


Dr. AlagozPrefix Averages (Linear) (*)

The following algorithm computes prefix averages in linear time by keeping a running sum

Algorithm prefixAverages2(X, n)Input array X of n integersOutput array A of prefix averages of X #operationsA new array of n integers ns 0 1for i 0 to n 1 do n

s s X[i] nA[i] s (i 1) n

return A 1Algorithm prefixAverages2 runs in O(n) time


Dr. AlagozThe Random Access Machine (RAM) Model (*)

A CPU

A potentially unbounded bank of memory cells, each of which can hold an arbitrary number or character

01

2

Memory cells are numbered and accessing any cell in memory takes unit time.


Dr. Alagoz

Recurrence Relations


Dr. Alagoz

dr. alagoz chapter 2 analysis of algorithms algorithm input output an algorithm is a step-by-step...

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